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756 Chapter 7 Systems of Equations and Inequalities
Preview Exercises
51. A modernistic painting consists of triangles, rectangles, and
pentagons, all drawn so as to not overlap or share sides.
Within each rectangle are drawn 2 red roses and each
pentagon contains 5 carnations. How many triangles,
rectangles, and pentagons appear in the painting if the painting contains a total of 40 geometric figures, 153 sides of
geometric figures, and 72 flowers?
Exercises 53–55 will help you prepare for the material covered in
the next section.
2
3
53. Subtract:
.
x - 4
x + 2
2x
5x - 3
54. Add:
+
.
2
x2 + 1
1x2 + 12
Group Exercise
55. Solve:
52. Group members should develop appropriate functions that
model each of the projections shown in Exercise 43.
Section
7.3
A + B = 3
c 2A - 2B + C = 17
4A - 2C = 14.
Partial Fractions
Objectives
� Decompose P , where Q has
Q
only distinct linear factors.
� Decompose P , where Q has
Q
repeated linear factors.
� Decompose P , where Q has
Q
a nonrepeated prime quadratic
factor.
� Decompose P , where Q has
Q
a prime, repeated quadratic
factor.
T
he rising and setting of the sun
suggest the obvious: Things
change over time. Calculus is
the study of rates of change,
allowing the motion of the
rising sun to be measured
by “freezing the frame” at
one instant in time. If you
are given a function, calculus
reveals its rate of change at any
“frozen” instant. In this section,
you will learn an algebraic technique
used in calculus to find a function if its rate of change is known. The technique
involves expressing a given function in terms of simpler functions.
The Idea behind Partial Fraction Decomposition
We know how to use common denominators to write a sum or difference of rational
expressions as a single rational expression. For example,
31x + 22 - 21x - 42
3
2
=
x - 4
x + 2
1x - 421x + 22
=
3x + 6 - 2x + 8
x + 14
=
.
1x - 421x + 22
1x - 421x + 22
For solving the kind of calculus problem described in the section opener, we must
reverse this process:
Partial fraction
x + 14
is
(x − 4)(x + 2)
expressed as the
sum of two
simpler fractions.
Partial fraction
x+14
3
–2
=
+
.
(x-4)(x+2)
x-4
x+2
This is the partial fraction
x + 14
decomposition of
.
(x − 4)(x + 2)
Each of the two fractions on the right is called a partial fraction. The sum of these
fractions is called the partial fraction decomposition of the rational expression on
the left-hand side.
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Section 7.3 Partial Fractions
757
Partial fraction decompositions can be written for rational expressions of the
P1x2
, where P and Q have no common factors and the highest power in the
form
Q1x2
numerator is less than the highest power in the denominator. In this section, we will
show you how to write the partial fraction decompositions for each of the following
rational expressions:
9x2-9x+6
(2x-1)(x+2)(x-2)
5x3-3x2+7x-3
.
(x2+1)2
P(x) = 9x2 − 9x + 6; highest power = 2
Q(x) = (2x − 1)(x + 2)(x − 2); multiplying factors,
highest power = 3.
P(x) = 5x3 − 3x2 + 7x − 3; highest power = 3
Q(x) = (x2 + 1)2; squaring the expression,
highest power = 4.
The partial fraction decomposition of a rational expression depends on the
factors of the denominator. We consider four cases involving different kinds of
factors in the denominator:
1.
2.
3.
4.
�
Decompose P , where Q has only
Q
distinct linear factors.
The denominator is a product of distinct linear factors.
The denominator is a product of linear factors, some of which are repeated.
The denominator has prime quadratic factors, none of which is repeated.
The denominator has a repeated prime quadratic factor.
The Partial Fraction Decomposition of a
Rational Expression with Distinct Linear Factors
in the Denominator
If the denominator of a rational expression has a linear factor of the form ax + b,
then the partial fraction decomposition will contain a term of the form
A
.
ax+b
Constant
Linear factor
Each distinct linear factor in the denominator produces a partial fraction of the
form constant over linear factor. For example,
B
C
9x2-9x+6
A
=
+
+
.
x+2
x-2
(2x-1)(x+2)(x-2)
2x-1
We write a constant over each linear factor
in the denominator.
P1x2
The Partial Fraction Decomposition of
: Q1x2 Has Distinct
Q1x2
Linear Factors
The form of the partial fraction decomposition for a rational expression with
distinct linear factors in the denominator is
P1x2
1a1x + b121a2x + b221a3x + b32 Á 1anx + bn2
=
A1
A2
A3
An
+
+
+ Á +
.
a1x + b1
a2x + b2
a3x + b3
anx + bn
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758 Chapter 7 Systems of Equations and Inequalities
EXAMPLE 1
Partial Fraction Decomposition with Distinct Linear
Factors
Find the partial fraction decomposition of
x + 14
.
1x - 421x + 22
Solution We begin by setting up the partial fraction decomposition with the
unknown constants. Write a constant over each of the two distinct linear factors in
the denominator.
x + 14
A
B
=
+
1x - 421x + 22
x - 4
x + 2
Our goal is to find A and B. We do this by multiplying both sides of the equation by
the least common denominator, 1x - 421x + 22.
(x-4)(x+2)
x+14
A
B
=(x-4)(x+2)a
+
b
(x-4)(x+2)
x-4
x+2
We use the distributive property on the right side.
1x - 42 1x + 22
x + 14
1x - 42 1x + 22
= 1x - 42 1x + 22
A
B
+ 1x - 42 1x + 22
1x - 42
1x + 22
Dividing out common factors in numerators and denominators, we obtain
x + 14 = A1x + 22 + B1x - 42.
To find values for A and B that make both sides equal, we’ll express the sides in
exactly the same form by writing the variable x-terms and then writing the constant
terms. Apply the distributive property on the right side.
x+14=Ax+2A+Bx-4B
Distribute A and B over the parentheses.
x+14=Ax+Bx+2A-4B
Rearrange terms.
1x+14=(A+B)x+(2A-4B)
Rewrite to identify the coefficient of x
and the constant term.
As shown by the arrows, if two polynomials are equal, coefficients of like powers of
x must be equal 1A + B = 12 and their constant terms must be equal
12A - 4B = 142. Consequently, A and B satisfy the following two equations:
b
A + B = 1
2A - 4B = 14.
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Section 7.3 Partial Fractions
759
We can use the addition method to solve this linear system in two variables. By
multiplying the first equation by -2 and adding equations, we obtain A = 3 and
B = - 2. Thus,
x + 14
A
B
3
-2
2
3
=
+
=
+
b.
aor
1x - 421x + 22
x - 4
x + 2
x - 4
x + 2
x - 4
x + 2
Steps in Partial Fraction Decomposition
1. Set up the partial fraction decomposition with the unknown constants
A, B, C, etc., in the numerators of the decomposition.
2. Multiply both sides of the resulting equation by the least common
denominator.
3. Simplify the right side of the equation.
4. Write both sides in descending powers, equate coefficients of like powers of
x, and equate constant terms.
5. Solve the resulting linear system for A, B, C, etc.
6. Substitute the values for A, B, C, etc., into the equation in step 1 and write
the partial fraction decomposition.
Study Tip
You will encounter some examples in which the denominator of the given rational expression
is not already factored. If necessary, begin by factoring the denominator. Then apply the six
steps needed to obtain the partial fraction decomposition.
Check Point
�
Decompose P , where Q has
Q
repeated linear factors.
1
Find the partial fraction decomposition of
5x - 1
.
1x - 321x + 42
The Partial Fraction Decomposition of a Rational
Expression with Linear Factors in the Denominator,
Some of Which Are Repeated
Suppose that 1ax + b2n is a factor of the denominator. This means that the linear
factor ax + b is repeated n times.When this occurs, the partial fraction decomposition
will contain a sum of n fractions for this factor of the denominator.
The Partial Fraction Decomposition of
P1x2
Q1x2
: Q1x2 Has Repeated
Linear Factors
The form of the partial fraction decomposition for a rational expression containing
the linear factor ax + b occurring n times as its denominator is
A3
P(x)
A1
A2
An
=
+
+
+. . .+
.
ax+b
(ax+b)2
(ax+b)3
(ax+b)n
(ax+b)n
Include one fraction with a constant numerator for each power of ax + b.
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760 Chapter 7 Systems of Equations and Inequalities
EXAMPLE 2
Partial Fraction Decomposition with Repeated
Linear Factors
Find the partial fraction decomposition of
x - 18
.
x1x - 322
Solution
Study Tip
Avoid this common error:
A
B
C
x-18
= +
+
x
x-3
x-3
x(x-3)2
Step 1 Set up the partial fraction decomposition with the unknown
constants. Because the linear factor x - 3 occurs twice, we must include one
fraction with a constant numerator for each power of x - 3.
B
A
C
x - 18
+
=
+
2
x
x
3
x1x - 32
1x - 322
Listing x − 3 twice does not
take into account (x − 3)2.
Step 2 Multiply both sides of the resulting equation by the least common
denominator. We clear fractions, multiplying both sides by x1x - 322, the least
common denominator.
x-18
B
C
2 A
x(x-3)2 B
+
R =x(x-3) B +
R
x(x-3)2
x
x-3
(x-3)2
We use the distributive property on the right side.
x 1x - 322
#
x - 18
A
B
C
= x 1x - 322 #
+ x1x - 322 #
+ x 1x - 322 #
2
x
1x
32
x 1x - 32
1x - 322
Dividing out common factors in numerators and denominators, we obtain
x - 18 = A1x - 322 + Bx1x - 32 + Cx.
Step 3 Simplify the right side of the equation. Square x - 3. Then apply the
distributive property.
x - 18 = A1x2 - 6x + 92 + Bx1x - 32 + Cx
Square x - 3 using
1A - B22 = A2 - 2AB + B2.
x - 18 = Ax2 - 6Ax + 9A + Bx2 - 3Bx + Cx
Apply the distributive property.
Step 4 Write both sides in descending powers, equate coefficients of like powers
of x, and equate constant terms. The left side, x - 18, is in descending powers of
x: x - 18x0. We will write the right side in descending powers of x.
x - 18 = Ax2 + Bx2 - 6Ax - 3Bx + Cx + 9A
Rearrange terms on the right side.
Express both sides in the same form.
0x2 + 1x - 18 = 1A + B2x2 + 1- 6A - 3B + C2x + 9A
Rewrite to identify
coefficients and the
constant term.
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Section 7.3 Partial Fractions
761
Equating coefficients of like powers of x and equating constant terms results in the
following system of linear equations:
A + B = 0
c - 6A - 3B + C = 1
9A = - 18.
Step 5 Solve the resulting system for A, B, and C. Dividing both sides of the last
equation by 9, we obtain A = - 2. Substituting -2 for A in the first equation,
A + B = 0, gives -2 + B = 0, so B = 2. We find C by substituting -2 for A and 2
for B in the middle equation, - 6A - 3B + C = 1. We obtain C = - 5.
Step 6 Substitute the values of A, B, and C, and write the partial fraction
decomposition. With A = - 2, B = 2, and C = - 5, the required partial fraction
decomposition is
B
2
x - 18
A
C
2
5
+
=
+
= - +
.
2
2
x
x
x
3
x
3
x1x - 32
1x - 32
1x - 322
Check Point
�
Decompose P , where Q has a
Q
nonrepeated prime quadratic
factor.
2
Find the partial fraction decomposition of
x + 2
.
x1x - 122
The Partial Fraction Decomposition of a Rational
Expression with Prime, Nonrepeated Quadratic Factors
in the Denominator
Our final two cases of partial fraction decomposition involve prime quadratic
factors of the form ax2 + bx + c. Based on our work with the discriminant, we
know that ax2 + bx + c is prime and cannot be factored over the integers if
b2 - 4ac 6 0 or if b2 - 4ac is not a perfect square.
P1x2
The Partial Fraction Decomposition of
: Q1x2 Has a Nonrepeated,
Q1x2
Prime Quadratic Factor
If ax2 + bx + c is a prime quadratic factor of Q1x2, the partial fraction
decomposition will contain a term of the form
Ax+B
.
ax2+bx+c
Linear numerator
Quadratic factor
The voice balloons in the box show that each distinct prime quadratic factor in
the denominator produces a partial fraction of the form linear numerator over
quadratic factor. For example,
Bx+C
3x2+17x+14
A
=
+ 2
.
x +2x+4
(x-2)(x2+2x+4)
x-2
We write a constant
over the linear factor
in the denominator.
We write a linear numerator
over the prime quadratic
factor in the denominator.
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762 Chapter 7 Systems of Equations and Inequalities
Our next example illustrates how a linear system in three variables is used to
determine values for A, B, and C.
EXAMPLE 3
Partial Fraction Decomposition
Find the partial fraction decomposition of
3x2 + 17x + 14
.
1x - 221x2 + 2x + 42
Solution
Step 1 Set up the partial fraction decomposition with the unknown constants. We
put a constant 1A2 over the linear factor and a linear expression 1Bx + C2 over the
prime quadratic factor.
3x2 + 17x + 14
A
Bx + C
=
+ 2
2
x - 2
1x - 221x + 2x + 42
x + 2x + 4
Step 2 Multiply both sides of the resulting equation by the least common
denominator. We clear fractions, multiplying both sides by (x - 2)(x2 + 2x + 4), the
least common denominator.
(x-2)(x2+2x+4) B
3x2+17x+14
A
Bx+C
2
+ 2
R=(x-2)(x +2x+4) B
R
2
(x-2)(x +2x+4)
x-2
x +2x+4
We use the distributive property on the right side.
1x - 22 1x2 + 2x + 42
#
3x2 + 17x + 14
1x - 22 1x2 + 2x + 42
= 1x - 22 1x2 + 2x + 42 #
A
Bx + C
+ 1x - 22 1x2 + 2x + 42 # 2
x - 2
x + 2x + 4
Dividing out common factors in numerators and denominators, we obtain
3x2 + 17x + 14 = A1x2 + 2x + 42 + 1Bx + C21x - 22.
Step 3 Simplify the right side of the equation. We simplify on the right side by
distributing A over each term in parentheses and multiplying 1Bx + C21x - 22
using the FOIL method.
3x2 + 17x + 14 = Ax2 + 2Ax + 4A + Bx2 - 2Bx + Cx - 2C
Step 4 Write both sides in descending powers, equate coefficients of like powers
of x, and equate constant terms. The left side, 3x2 + 17x + 14, is in descending
powers of x. We write the right side in descending powers of x
3x 2 + 17x + 14 = Ax2 + Bx2 + 2Ax - 2Bx + Cx + 4A - 2C
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Section 7.3 Partial Fractions
763
and express both sides in the same form.
3x2 + 17x + 14 = 1A + B2x2 + 12A - 2B + C2x + 14A - 2C2
Equating coefficients of like powers of x and equating constant terms results in the
following system of linear equations:
A + B = 3
c 2A - 2B + C = 17
4A - 2C = 14.
Step 5 Solve the resulting system for A, B, and C. Because the first equation
involves A and B, we can obtain another equation in A and B by eliminating C from
the second and third equations. Multiply the second equation by 2 and add
equations. Solving in this manner, we obtain A = 5, B = - 2, and C = 3.
Step 6 Substitute the values of A, B, and C, and write the partial fraction
decomposition. With A = 5, B = - 2, and C = 3, the required partial fraction
decomposition is
3x2 + 17x + 14
A
Bx + C
5
-2x + 3
=
+ 2
=
+ 2
.
x - 2
x - 2
1x - 221x2 + 2x + 42
x + 2x + 4
x + 2x + 4
Technology
Numeric Connections
You can use the 冷 TABLE 冷 feature of a graphing
utility to check a partial fraction decomposition.
To check the result of Example 3, enter the given
rational function and its partial fraction
decomposition:
y1 =
3x 2 + 17x + 14
1x - 221x2 + 2x + 42
5
-2x + 3
y2 =
.
+ 2
x - 2
x + 2x + 4
Check Point
3
No matter how far up or
down we scroll, y1 = y2 , so
the decomposition appears
to be correct.
Find the partial fraction decomposition of
8x2 + 12x - 20
.
1x + 321x2 + x + 22
�
Decompose P , where Q has a
Q
prime, repeated quadratic factor.
The Partial Fraction Decomposition of a Rational
Expression with a Prime, Repeated Quadratic Factor in
the Denominator
Suppose that 1ax2 + bx + c2 is a factor of the denominator and that ax2 + bx + c
cannot be factored further. This means that the quadratic factor ax2 + bx + c
occurs n times. When this occurs, the partial fraction decomposition will contain a
linear numerator for each power of ax 2 + bx + c.
n
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764 Chapter 7 Systems of Equations and Inequalities
P1x2
The Partial Fraction Decomposition of
: Q1x2 Has a Prime,
Q1x2
Repeated Quadratic Factor
The form of the partial fraction decomposition for a rational expression containing
the prime factor ax 2 + bx + c occurring n times as its denominator is
P(x)
A1x+B1
A2x+B2
A3x+B3
Anx+Bn
. . .+
+ 2
.
n=
2
2+
2
3+
(ax +bx+c)
ax +bx+c
(ax +bx+c)
(ax +bx+c)
(ax2+bx+c)n
2
Include one fraction with a linear numerator for each power of ax2 + bx + c.
EXAMPLE 4
Study Tip
When the denominator of a rational
expression contains a power of a
linear factor, set up the partial
fraction decomposition with constant
numerators (A, B, C, etc.). When the
denominator of a rational expression
contains a power of a prime quadratic
factor, set up the partial fraction
decomposition with linear numerators (Ax + B, Cx + D, etc.).
Partial Fraction Decomposition with a Repeated
Quadratic Factor
Find the partial fraction decomposition of
5x 3 - 3x2 + 7x - 3
1x2 + 12
2
Solution
.
Step 1 Set up the partial fraction decomposition with the unknown
constants. Because the quadratic factor x2 + 1 occurs twice, we must include one
fraction with a linear numerator for each power of x2 + 1.
5x3 - 3x2 + 7x - 3
Ax + B
Cx + D
+
2
2
x + 1
1x + 12
1x2 + 12
Step 2 Multiply both sides of the resulting equation by the least common
2
denominator. We clear fractions, multiplying both sides by 1x2 + 12 , the least
common denominator.
2
(x2+1)2 B
2
=
5x3-3x2+7x-3
Cx+D
2
2 Ax+B
+ 2
R=(x +1) B 2
R
2
2
(x +1)
x +1
(x +1)2
Now we multiply and simplify.
5x3 - 3x2 + 7x - 3 = 1x2 + 121Ax + B2 + Cx + D
Step 3 Simplify the right side of the equation. We multiply 1x2 + 121Ax + B2
using the FOIL method.
5x3 - 3x2 + 7x - 3 = Ax3 + Bx2 + Ax + B + Cx + D
Step 4 Write both sides in descending powers, equate coefficients of like powers
of x, and equate constant terms.
5x3 - 3x2 + 7x - 3 = Ax3 + Bx2 + Ax + Cx + B + D
5x3-3x2+7x-3=Ax3+Bx2+(A+C)x+(B+D)
Equating coefficients of like powers of x and equating constant terms results in the
following system of linear equations:
A
B
d
A + C
B + D
= 5
= -3
= 7
= - 3.
With A = 5, we immediately obtain C = 2.
With B = - 3, we immediately obtain D = 0.
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Section 7.3 Partial Fractions
765
Step 5 Solve the resulting system for A, B, C, and D. Based on our observations
in step 4, A = 5, B = - 3, C = 2, and D = 0.
Step 6 Substitute the values of A, B, C, and D, and write the partial fraction
decomposition.
5x3 - 3x2 + 7x - 3
1x + 12
Check Point
=
2
2
4
Ax + B
5x - 3
Cx + D
2x
= 2
+
+
2
2
2
2
2
x + 1
x + 1
1x + 12
1x + 12
Find the partial fraction decomposition of
2x 3 + x + 3
1x2 + 12
2
.
Study Tip
When a rational expression contains a power of a factor in the denominator, be sure to set up
the partial fraction decomposition to allow for every natural-number power of that factor less
than or equal to the power. Example:
2x + 1
1x - 522x3
=
A
C
E
B
D
+
+
+ 2 + 3
x - 5
x
(x - 5)2
x
x
Although (x − 5)2 and x2 are quadratic,
they are still expressed as powers
of linear factors, x − 5 and x.
Thus, the numerators are constant.
Exercise Set 7.3
Practice Exercises
In Exercises 1–8, write the form of the partial fraction
decomposition of the rational expression. It is not necessary to
solve for the constants.
13.
7x - 4
x2 - x - 12
14.
9x + 21
x2 + 2x - 15
15.
4
2x - 5x - 3
16.
x
x + 2x - 3
2
2
1.
11x - 10
1x - 221x + 12
2.
5x + 7
1x - 121x + 32
17.
18.
4x2 - 5x - 15
x1x + 121x - 52
3.
6x2 - 14x - 27
1x + 221x - 322
4x2 + 13x - 9
x1x - 121x + 32
4.
3x + 16
1x + 121x - 222
5x2 - 6x + 7
1x - 121x2 + 12
19.
4x 2 - 7x - 3
x3 - x
20.
2x2 - 18x - 12
x3 - 4x
6.
21.
6x - 11
1x - 122
22.
x
1x + 122
23.
x2 - 6x + 3
1x - 223
24.
2x2 + 8x + 3
1x + 123
25.
x2 + 2x + 7
x1x - 122
26.
3x2 + 49
x1x + 722
27.
x2
1x - 1221x + 12
28.
x2
1x - 1221x + 122
5.
3
7.
x + x
2
2
1x + 42
2
5x2 - 9x + 19
1x - 421x2 + 52
2
8.
7x - 9x + 3
1x2 + 72
2
In Exercises 9–42, write the partial fraction decomposition of each
rational expression.
x
9.
1x - 321x - 22
11.
3x + 50
1x - 921x + 22
1
10.
x1x - 12
12.
5x - 1
1x - 221x + 12
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766 Chapter 7 Systems of Equations and Inequalities
5x2 - 6x + 7
1x - 121x2 + 12
30.
31.
5x2 + 6x + 3
1x + 121x2 + 2x + 22
33.
35.
29.
37.
39.
41.
5x2 - 9x + 19
1x - 421x2 + 52
Writing in Mathematics
32.
9x + 2
1x - 221x2 + 2x + 22
54. Explain how to find the partial fraction decomposition of a
rational expression with distinct linear factors in the
denominator.
x + 4
x 1x2 + 42
34.
10x2 + 2x
1x - 1221x2 + 22
55. Explain how to find the partial fraction decomposition of
a rational expression with a repeated linear factor in the
denominator.
6x2 - x + 1
x3 + x2 + x + 1
36.
3x2 - 2x + 8
x3 + 2x2 + 4x + 8
56. Explain how to find the partial fraction decomposition of
a rational expression with a prime quadratic factor in the
denominator.
x2 + 2x + 3
57. Explain how to find the partial fraction decomposition of a
rational expression with a repeated, prime quadratic factor in
the denominator.
2
x3 + x2 + 2
38.
1x + 22
2
2
x3 - 4x2 + 9x - 5
40.
1x - 2x + 32
2
2
4x2 + 3x + 14
x3 - 8
42.
53. Explain what is meant by the partial fraction decomposition
of a rational expression.
1x2 + 42
2
3x3 - 6x2 + 7x - 2
1x - 2x + 22
2
2
3x - 5
x3 - 1
58. How can you verify your result for the partial fraction
decomposition for a given rational expression without using
a graphing utility?
Technology Exercise
59. Use the 冷 TABLE 冷 feature of a graphing utility to verify
any three of the decompositions that you obtained in
Exercises 9–42.
Practice Plus
In Exercises 43–46, perform each long division and write the
partial fraction decomposition of the remainder term.
43.
x5 + 2
x2 - 1
44.
x5
x - 4x + 4
45.
x4 - x2 + 2
x3 - x2
46.
x4 + 2x3 - 4x2 + x - 3
x2 - x - 2
2
In Exercises 47–50, write the partial fraction decomposition of
each rational expression.
1
47. 2
x - c2
1c Z 02
ax + b
48. 2
x - c2
1c Z 02
1c Z 02
49.
ax + b
1x - c22
50.
1
x2 - ax - bx + ab
1a Z b2
2
52. Find the partial fraction decomposition for
and use
x1x + 22
the result to find the following sum:
2
3#5
61. I apply partial fraction decompositions for rational expresP1x2
sions of the form
, where P and Q have no common
Q1x2
factors and the degree of P is greater than the degree of Q.
63. Because 1x + 322 consists of two factors of x + 3, I set up the
following partial fraction decomposition:
1
1
1
1
+ # + # + Á +
.
1#2
2 3
3 4
99 # 100
1#3
60. Partial fraction decomposition involves finding a single
rational expression for a given sum or difference of rational
expressions.
A
7x2 + 9x + 3
Bx + C
=
.
+ 2
x + 5
1x + 521x2 - 3x + 22
x - 3x + 2
1
51. Find the partial fraction decomposition for
and use
x1x + 12
the result to find the following sum:
+
Make Sense? In Exercises 60–63, determine whether each
statement makes sense or does not make sense, and explain
your reasoning.
62. Because x + 5 is linear and x2 - 3x + 2 is quadratic, I set
up the following partial fraction decomposition:
Application Exercises
2
Critical Thinking Exercises
2
+
5#7
+ Á +
2
.
99 # 101
5x + 2
A
B
=
+
.
x + 3
x + 3
1x + 322
64. Use an extension of the Study Tip on page 764 to describe
how to set up the partial fraction decomposition of a rational
expression that contains powers of a prime cubic factor in the
denominator. Give an example of such a decomposition.
65. Find the partial fraction decomposition of
4x 2 + 5x - 9
.
x3 - 6x - 9
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Page 767
Section 7.4 Systems of Nonlinear Equations in Two Variables
Preview Exercises
Exercises 66–68 will help you prepare for the material covered in
the next section.
66. Solve by the substitution method:
b
Section
4x + 3y = 4
y = 2x - 7.
7.4
Objectives
� Recognize systems of
�
�
�
�
nonlinear equations in two
variables.
Solve nonlinear systems by
substitution.
Solve nonlinear systems by
addition.
Solve problems using
systems of nonlinear
equations.
Recognize systems of nonlinear
equations in two variables.
767
67. Solve by the addition method:
2x + 4y = - 4
b
3x + 5y = - 3.
68. Graph x - y = 3 and 1x - 222 + 1y + 322 = 4 in the same
rectangular coordinate system. What are the two intersection
points? Show that each of these ordered pairs satisfies both
equations.
Systems of Nonlinear Equations in Two Variables
S
cientists debate the probability that a “doomsday
rock” will collide with Earth.
It has been estimated that an
asteroid, a tiny planet that
revolves around the sun,
crashes into Earth about once
every 250,000 years, and that
such a collision would have
disastrous results. In 1908, a small fragment
struck Siberia, leveling thousands of acres of trees.
One theory about the extinction of dinosaurs
65 million years ago involves Earth’s collision
with a large asteroid and the resulting drastic
changes in Earth’s climate.
Understanding the path of Earth and the path
of a comet is essential to detecting threatening space debris. Orbits about the sun
are not described by linear equations in the form Ax + By = C. The ability to solve
systems that contain nonlinear equations provides NASA scientists watching for
troublesome asteroids with a way to locate possible collision points with
Earth’s orbit.
Systems of Nonlinear Equations and Their Solutions
A system of two nonlinear equations in two variables, also called a nonlinear
system, contains at least one equation that cannot be expressed in the form
Ax + By = C. Here are two examples:
e
x2=2y+10
3x-y=9
Not in the form
Ax + By = C.
The term x2 is
not linear.
e
y=x2+3
x2+y2=9.
Neither equation is in
the form Ax + By = C.
The terms x2 and y2 are
not linear.
A solution of a nonlinear system in two variables is an ordered pair of real
numbers that satisfies both equations in the system. The solution set of the system is
the set of all such ordered pairs. As with linear systems in two variables, the solution
of a nonlinear system (if there is one) corresponds to the intersection point(s) of the
graphs of the equations in the system. Unlike linear systems, the graphs can be
circles, parabolas, or anything other than two lines. We will solve nonlinear systems
using the substitution method and the addition method.